如何在MSSQL 2005中创建递归查询?

在SQL Server 2005上,您可以使用 CTEs (Common Table Expressions) :

with Hierachy(CustomerID, ParentID, Name, Level)
as
(
select CustomerID, ParentID, Name, 0 as Level
    from Customers c
    where c.CustomerID = 2 -- insert parameter here
    union all
    select c.CustomerID, c.ParentID, c.Name, ch.Level + 1
    from Customers c
    inner join Hierachy ch
    on c.ParentId = ch.CustomerID
)
select CustomerID, ParentID, Name
from Hierachy
where Level > 0

对于自下而上,使用mathieu的答案,稍作修改:

with Hierachy(CustomerID, ParentID, Name, Level)
as
(
select CustomerID, ParentID, Name, 0 as Level
    from Customers c
    where c.CustomerID = 2 -- insert parameter here
    union all
    select c.CustomerID, c.ParentID, c.Name, ch.Level + 1
    from Customers c
    inner join Hierachy ch

    -- EDITED HERE --
    on ch.ParentId = c.CustomerID
    ----------------- 

)
select CustomerID, ParentID, Name
from Hierachy
where Level > 0

没有存储过程,你就无法在SQL中进行递归。解决这个问题的方法是使用嵌套集,它们基本上将SQL中的树建模为集。

请注意,这将需要更改当前数据模型,或者可能需要弄清楚如何在原始模型上创建视图。

Postgresql示例(使用很少的Postgresql扩展,只有SERIAL和ON COMMIT DROP,大多数RDBMS将具有类似的功能):

设置:

CREATE TABLE objects(
    id SERIAL PRIMARY KEY,
    name TEXT,
    lft INT,
    rgt INT
);

INSERT INTO objects(name, lft, rgt) VALUES('The root of the tree', 1, 2);

添加孩子:

START TRANSACTION;

-- postgresql doesn't support variables so we create a temporary table that 
-- gets deleted after the transaction has finished.

CREATE TEMP TABLE left_tmp(
    lft INT
) ON COMMIT DROP; -- not standard sql

-- store the left of the parent for later use
INSERT INTO left_tmp (lft) VALUES((SELECT lft FROM objects WHERE name = 'The parent of the newly inserted node'));

-- move all the children already in the set to the right
-- to make room for the new child
UPDATE objects SET rgt = rgt + 2 WHERE rgt > (SELECT lft FROM left_tmp LIMIT 1);
UPDATE objects SET lft = lft + 2 WHERE lft > (SELECT lft FROM left_tmp LIMIT 1);

-- insert the new child
INSERT INTO objects(name, lft, rgt) VALUES(
    'The name of the newly inserted node', 
    (SELECT lft + 1 FROM left_tmp LIMIT 1), 
    (SELECT lft + 2 FROM left_tmp LIMIT 1)
);

COMMIT;

从下到上显示轨迹:

SELECT
    parent.id, parent.lft
FROM
    objects AS current_node
INNER JOIN
    objects AS parent
ON
    current_node.lft BETWEEN parent.lft AND parent.rgt
WHERE
    current_node.name = 'The name of the deepest child'
ORDER BY
    parent.lft;

显示整个树:

SELECT
    REPEAT('   ', CAST((COUNT(parent.id) - 1) AS INT)) || '- ' || current_node.name AS indented_name
FROM
    objects current_node
INNER JOIN
    objects parent
ON
    current_node.lft BETWEEN parent.lft AND parent.rgt
GROUP BY
    current_node.name,
    current_node.lft
ORDER BY
    current_node.lft;

从树的某个元素中选择所有内容:

SELECT
    current_node.name AS node_name
FROM
    objects current_node
INNER JOIN
    objects parent
ON
    current_node.lft BETWEEN parent.lft AND parent.rgt
AND
    parent.name = 'child'
GROUP BY
    current_node.name,
    current_node.lft
ORDER BY
    current_node.lft;

除非我遗漏了什么,否则递归是不必要的。..

SELECT d.NAME FROM Customers As d
INNER JOIN Customers As p ON p.CustomerID = d.ParentID
WHERE p.Name = 'James'